Proof.

Let a,b∈ℚ with φ⁢(m)=a+b⁢n. Since φ⁢(a)=a and φ is injective, b≠0. Also, m=φ⁢(m)=φ⁢((m)2)=(φ⁢(m))2=(a+b⁢n)2=a2+2⁢a⁢b⁢n+b2⁢n. If a≠0, then n=m-a2-b2⁢n2⁢a⁢b∈ℚ, a contradiction. Thus, a=0. Therefore, m=b2⁢n. Since m is squarefree, b2=1. Hence, m=n, a contradiction. It follows that K and L are not isomorphic.
∎

Corollary.

Proof.

Note that there are infinitely many elements of S. Moreover, if m and n are distinct elements of S, then ℚ⁢(m) and ℚ⁢(n) are not isomorphic and thus cannot be equal.
∎

Note that the above corollary could have also been obtained by using the result regarding Galois groups of finite abelian extensions of ℚ (http://planetmath.org/GaloisGroupsOfFiniteAbelianExtensionsOfMathbbQ). On the other hand, using this result to prove the above corollary can be likened to “using a sledgehammer to kill a housefly”.